elif using dictionary but without excecute the values

Question

Hi let's say that I want to replace a large el-ifs statement by a dictionary. A simplify version of this is as follows:

def functionA(a1, a2, a3):

    results = {
        'case1': a2/a3,
        'case2': 1,
        ...
    }
    return results[a1]

so a1 would be a string ('case1' or 'case2' or ...) the problem is in some cases the a3 it maybe 0 so the results dictionary could not be define (in all those cases a1 would not be 'case1'). For instance:

functionA('case2', 1.0, 3.0)
Out[81]: 1
functionA('case2', 1.0, 0.0)
ZeroDivisionError: integer division or modulo by zero

So in the second case I expected 1 but I am getting an error.

A possible solution is:

def functionA(a1, a2, a3):
    results = {
        'case1': str(a2) + '/' + str(a3),
        'case2': str(1),
    }

    return eval(results[a1])

Since I have a multiple cases with complex calculations, is there any better solution?

that's a typical case for default parameters. `a3` should have a default, so you don't have to pass 0 when you only need 1 param. — Jean-François Fabre, Oct 26 '17 at 17:22
Or you could use lambda expressions instead of relying on string expressions being evaluated. — Martijn Pieters, Oct 26 '17 at 17:35

score 4 · Accepted Answer · edited Oct 26 '17 at 21:14

def functionA(a1, a2, a3):
    results = {
        'case1': a2 / a3 if a3 != 0.0 else a2,
        'case2': 1,
        # ...
    }
    return results[a1]

However, I would advise against this approach as the entire dictionary has to be computed first only to pick up a single value.

Instead, if you really want to use dictionaries and have it also efficient, I would suggest:

myproc = {
    'case1': lambda a2, a3: a2 / a3 if a3 != 0.0 else a2,
    'case2': lambda a2, a3: 1
}

Then:

>>> myproc['case1'](2, 3)
0.6666666666666666

score 1 · Answer 2 · answered Oct 26 '17 at 17:21

1

You can use ternary operator inside your dictionary

def functionA(a1, a2, a3):

    results = {
        'case1': a2 if a3==0 else a2/a3,
        'case2': 1
    }
    return results[a1]

Link : https://repl.it/NRkd

answered Oct 26 '17 at 17:21

Rahul Verma

2,946
14
27

1

To the `downvoter`: If you think there is any issue then you should come upfront and explain. – Rahul Verma Oct 26 '17 at 18:14
I do not think there is anything *fundamentally* wrong with your solution - so +1. – AGN Gazer Oct 26 '17 at 18:25

utengr · Answer 3 · 2017-10-26T17:23:32.563

0

If it's just that one value, you can do it with a simple if-else.

def functionA(a1, a2, a3):

     if a3:
        case1_value = a2/a3
     else: 
        case1_value = a2

    results = {
        'case1': case1_value,
        'case2': 1,
        ...
    }
    return results[a1]

Or create a dvision function with this if-else condition and use it in your dict value.

edited Oct 26 '17 at 17:23

answered Oct 26 '17 at 17:18

utengr

3,225
3
29
68

elif using dictionary but without excecute the values

3 Answers3